Import Question JSON

Current Question (ID: 21004)

Question:

$\text{Match List-I with List-II:}$ $\begin{array}{|c|c|} \hline \text{List-I} & \text{List-II} \\ \text{(Expressions)} & \text{(Property)} \\ \hline \Psi_{MO} = \Psi_A - \Psi_B & \text{I. Dipole moment} \\ \mu = Q \times r & \text{II. Bonding molecular orbital} \\ \frac{N_b - N_a}{2} & \text{III. Anti-bonding molecular orbital} \\ \Psi_{MO} = \Psi_A + \Psi_B & \text{IV. Bond order} \\ \hline \end{array}$ $\text{Choose the correct answer from the options given below:}$ $1. \text{A-II, B-I, C-IV, D-III}$ $2. \text{A-III, B-IV, C-I, D-II}$ $3. \text{A-III, B-I, C-IV, D-II}$ $4. \text{A-II, B-IV, C-I, D-III}$

Options:

1. $\text{A-II, B-I, C-IV, D-III}$
2. $\text{A-III, B-IV, C-I, D-II}$
3. $\text{A-III, B-I, C-IV, D-II}$
4. $\text{A-II, B-IV, C-I, D-III}$

Solution:

$\text{(1) Antibonding orbitals are formed through destructive interference of atomic orbitals, meaning they have a node or point of zero electron density between the nuclei.}$ $\Psi_A - \Psi_B = \Psi_{MO} \text{ is anti-bonding molecular orbital}$ $\text{(2) Dipole moment is calculated as the product of the magnitude of the charge (Q) and the distance (d) between the centers of positive and negative charges. Mathematically, it's represented as:}$ $\mu = Q \times r$ $\text{(3) Bond Order Formula can be defined as half of the difference between the number of electrons in bonding orbitals and antibonding orbitals}$ $\frac{N_b - N_a}{2} = \text{bond order}$ $\text{(4) Bonding molecular orbitals are formed when atomic orbitals combine in a way that their lobes overlap in phase (constructive interference), leading to a net increase in electron density between the nuclei.}$ $\Psi_A + \Psi_B = \Psi_{MO} \text{ is bonding molecular orbital.}$ $\text{Hence, option 3 is the correct answer.}$

Import JSON File

Upload a JSON file containing LaTeX/MathJax formatted question, options, and solution.

Expected JSON Format:

{
  "question": "The mass of carbon present in 0.5 mole of $\\mathrm{K}_4[\\mathrm{Fe(CN)}_6]$ is:",
  "options": [
    {
      "id": 1,
      "text": "1.8 g"
    },
    {
      "id": 2,
      "text": "18 g"
    },
    {
      "id": 3,
      "text": "3.6 g"
    },
    {
      "id": 4,
      "text": "36 g"
    }
  ],
  "solution": "\\begin{align}\n&\\text{Hint: Mole concept}\\\\\n&1 \\text{ mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6] = 6 \\text{ moles of carbon atom}\\\\\n&0.5 \\text{ mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6] = 6 \\times 0.5 \\text{ mol} = 3 \\text{ mol}\\\\\n&1 \\text{ mol of carbon} = 12 \\text{ g}\\\\\n&3 \\text{ mol carbon} = 12 \\times 3 = 36 \\text{ g}\\\\\n&\\text{Hence, 36 g mass of carbon present in 0.5 mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6].\n\\end{align}",
  "correct_answer": 4
}